The statement makes no sense if a person has 0 friends, so assume that everybody has at least one friend.  Also, assume friendship is reflexive (if I am your friend, then you are my friend).  

Consider n = 3.

Since we have assumed that everybody has at least one friend, there must be 2 or 3 friendships.

If there are three friendships, then everybody has two friends.

But if there are only two friendship, then the two less popular kids have only one friendship, which is less than the 2 friendships which their only friend have.  So 2/3 of the people have fewer friends than their friends.

So this strange statement seems possible.

Let's do some math, still for n = 3.

Let the probability that any pair of people are friends = p.

The prob(3 friendships|2 or 3 friendships) = p^3/(p^3 + 2(1-p)*p^2) = p/(3-2p)

The prob(2 friendships|2 or 3 friendships) = 1- (p/(3-2p)) = (3-3p)/(3-2p)

The probability that any given person has fewer friends than their friends =

    0 * prob(3 friendships|2 or 3 friendhips)

  + 2/3 * prob(2 friendships|2 or 3 friendhips)

  

  = (2-2p)/(3-2p).

  

This is greater than 1/2 if 2*(2-2p) > (3 - 2p) 

In other words, if p < 1/2.

So, the statement is not necessarily true.  It depends on how friendly people are.

As n increases, I expect that the threshold value of p increases, so the statement is more and more likely to be true.

Plus, as n increases, it makes sense that the actual value of p will decrease, so the statement is also more likely to be true.